In SHM with amplitude a, the potential energy and kinetic energy are equal to each other as displacement

Options

A. a. `a//sqrt2`
B. b. a/4
C. c. a/3
D. d. a/2

Explanation

To solve the problem, we need to find the displacement in Simple Harmonic Motion (SHM) when the potential energy (U) and kinetic energy (K) are equal.

### Step-by-Step Solution:

1. **Understanding SHM**:
In SHM, the displacement \( y \) can be expressed as:
\[ y = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Kinetic Energy (K)**:
The kinetic energy in SHM is given by:
\[ K = \frac{1}{2} m v^2 \]
where \( v \) is the velocity. The velocity can be derived from the displacement:
\[ v = \frac{dy}{dt} = A \omega \cos(\omega t) \]
Therefore, the kinetic energy becomes:
\[ K = \frac{1}{2} m (A \omega \cos(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \]

3. **Potential Energy (U)**:
The potential energy in SHM is given by:
\[ U = \frac{1}{2} m \omega^2 y^2 \]
Substituting \( y = A \sin(\omega t) \):
\[ U = \frac{1}{2} m \omega^2 (A \sin(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t) \]

4. **Setting Kinetic Energy Equal to Potential Energy**:
We need to find the displacement when \( K = U \):
\[ \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t) \]
Canceling the common terms:
\[ \cos^2(\omega t) = \sin^2(\omega t) \]

5. **Using Trigonometric Identity**:
From the equation \( \cos^2(\omega t) = \sin^2(\omega t) \), we can use the identity:
\[ \tan(\omega t) = 1 \]
This implies:
\[ \omega t = 45^\circ \quad \text{or} \quad \omega t = \frac{\pi}{4} \text{ radians} \]

6. **Finding Displacement**:
Now, substituting \( \omega t = \frac{\pi}{4} \) back into the displacement equation:
\[ y = A \sin\left(\frac{\pi}{4}\right) = A \cdot \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}} \]

### Final Answer:
The displacement when the potential energy and kinetic energy are equal is:
\[ y = \frac{A}{\sqrt{2}} \]